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\title{第三次作业(5-8)}
\author{付临\\3200104960\\信息与计算科学}
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\begin{document}
\maketitle
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\section{}
{\bfseries Solution:}\\
$\bullet$ The recursive definition of B-splines and the hat function are
\begin{align*}
    B_{i}^{n+1}(x)=\frac{x-t_{i-1}}{t_{i+n}-t_{i-1}}B_{i}^{n}(x)+
    \frac{t_{i+n+1}-x}{t_{i+n+1}-t_{i}}B_{i+1}^{n}(x)
\end{align*}
\begin{equation*}
    \hat{B_{i}}(x)= \left\{\begin{aligned}
        &\frac{x-t_{i-1}}{t_{i}-t_{i-1}} &x\in(t_{i-1},t_{i}]\\
        &\frac{t_{i+1}-x}{t_{i+1}-t_{i}} &x\in(t_{i},t_{i+1}]\\
        &0 &otherwise\\
    \end{aligned}\right.
\end{equation*}
Let n=1, we have
\begin{equation}
    B_{i}^{2}(x)=\frac{x-t_{i-1}}{t_{i+1}-t_{i-1}}B_{i}^{1}(x)+
    \frac{t_{i+2}-x}{t_{i+2}-t_{i}}B_{i+1}^{1}(x)
    \label{eq::f1}
\end{equation}
Because $B_{i}^{1}=\hat{B_{i}}$, so we have
\begin{equation}
    B_{i}^{1}=\hat{B_{i}}(x)= \left\{\begin{aligned}
        &\frac{x-t_{i-1}}{t_{i}-t_{i-1}} &x\in(t_{i-1},t_{i}]\\
        &\frac{t_{i+1}-x}{t_{i+1}-t_{i}} &x\in(t_{i},t_{i+1}]\\
        &0 &otherwise\\
    \end{aligned}\right.\label{eq::f2}
\end{equation}
\begin{equation}
    B_{i+1}^{1}=\hat{B_{i+1}}(x)= \left\{\begin{aligned}
        &\frac{x-t_{i}}{t_{i+1}-t_{i}} &x\in(t_{i},t_{i+1}]\\
        &\frac{t_{i+2}-x}{t_{i+2}-t_{i+1}} &x\in(t_{i+1},t_{i+2}]\\
        &0 &otherwise\\
    \end{aligned}\right.\label{eq::f3}
\end{equation}
So, plug the formula \ref{eq::f2} and the formula \ref{eq::f3} into the formula \ref{eq::f1}. 
We can get
\begin{equation*}
    B_{i}^{2}(x)=\left\{\begin{aligned}
        &\frac{(x-t_{i-1})^{2}}{(t_{i+1}-t_{i-1})(t_{i}-t_{i-1})} &x\in(t_{i-1},t_{i}]\\
        &\frac{(x-t_{i-1})(t_{i+1}-x)}{(t_{i+1}-t_{i-1})(t_{i+1}-t_{i})}+
        \frac{(t_{i+2}-x)(x-t_{i})}{(t_{i+2}-t_{i})(t_{i+1}-t_{i})} &x\in(t_{i},t_{i+1}]\\
        &\frac{(t_{i+2}-x)^2}{(t_{i+2}-t_{i})(t_{i+2}-t_{i+1})} &x\in(t_{i+1},t_{i+2}]\\
        &0 &otherwise\\
    \end{aligned}\right.
\end{equation*}

$\bullet$ When $x\in (t_{i-1},t_{i}]$
\begin{equation*}
    \frac{d}{dx}B_{i}^{2}(x)=\frac{2(x-t_{i-1})}{(t_{i+1}-t_{i-1})(t_{i}-t_{i-1})}
\end{equation*}
So $\frac{d}{dx}B_{i}^{2}(t_{i})=\frac{2}{t_{i+1}-t_{i-1}}$. Because 
\begin{align*}
    \lim_{h \to 0} \frac{B_{i}^{2}(t_{i}+h)-B_{i}^{2}(t_{i})}{h}&=\lim_{h \to 0} 
    \frac{\frac{(t_{i}-t_{i-1}+h)(t_{i+1}-t_{i}-h)}{(t_{i+1}-t_{i-1})(t_{i+1}-t_{i})}
    +\frac{(t_{i+2}-t_{i}-h)h}{(t_{i+2}-t_{i})(t_{i+1}-t_{i})}
    -\frac{t_{i}-t_{i-1}}{t_{i+1}-t_{i-1}}}{h}\\
    &=-\frac{t_{i}-t_{i-1}}{(t_{i+1}-t_{i-1})(t_{i+1}-t_{i})}+\frac{1}{t_{i+1}-t_{i-1}}
    +\frac{1}{t_{i+1}-t_{i}}\\
    &=\frac{2}{t_{i+1}-t_{i-1}}\\
    &=\frac{d}{dx}B_{i}^{2}(t_{i})
\end{align*}
So, $\frac{d}{dx}B_{i}^{2}(t_{i})$ is continuous at $t_{i}$.
\\When $x\in (t_{i},t_{i+1}]$
\begin{equation*}
    \frac{d}{dx}B_{i}^{2}(x)=\frac{-2x+t_{i+1}+t_{i-1}}{(t_{i+1}-t_{i-1})(t_{i+1}-t_{i})}
    +\frac{-2x+t_{i}+t_{i+2}}{(t_{i+2}-t_{i})(t_{i+1}-t_{i})}\\
\end{equation*}
So $\frac{d}{dx}B_{i}^{2}(t_{i+1})=\frac{2}{t_{i}-t_{i+2}}$. Because
\begin{align*}
    \lim_{h \to 0} \frac{B_{i}^{2}(t_{i+1}+h)-B_{i}^{2}(t_{i+1})}{h}&=
    \lim_{h \to 0} \frac{\frac{(t_{i+2}-t_{i+1}-h)^{2}}{(t_{i+2}-t_{i})(t_{i+2}-t_{i+1})}
    -\frac{t_{i+2}-t_{i-1}}{t_{i+2}-t_{i}}}{h}\\
    &=\frac{2}{t_{i}-t_{i+2}}\\
    &=\frac{d}{dx}B_{i}^{2}(x_{i+1})
\end{align*}
So, $\frac{d}{dx}B_{i}^{2}(x_{i+1})$ is continuous at $t_{i+1}$.

$\bullet$ When $x \in (t_{i-1},t_{i+1})$, we have
\begin{equation*}
    \frac{d}{dx}B_{i}^{2}(x)=\left\{\begin{aligned}
        &=\frac{2(x-t_{i-1})}{(t_{i+1}-t_{i-1})(t_{i}-t_{i-1})} &x \in (t_{i-1},t_{i}]\\
        &=\frac{-2x+t_{i+1}+t_{i-1}}{(t_{i+1}-t_{i-1})(t_{i+1}-t_{i})}
        +\frac{-2x+t_{i}+t_{i+2}}{(t_{i+2}-t_{i})(t_{i+1}-t_{i})} &x \in (t_{i},t_{i+1})\\
    \end{aligned}\right.
\end{equation*}
Obviously, when $x^* \in (t_{i-1},t_{i}]$, $\frac{d}{dx}B_{i}^{2}(x^*)\neq 0$. When 
$x^*\in (t_{i},t_{i+1})$, let $\frac{d}{dx}B_{i}^{2}(x^*)=0$. Then
\begin{align*}
    \frac{d}{dx}B_{i}^{2}(x^*)=\frac{-2x^*+t_{i+1}+t_{i-1}}{(t_{i+1}-t_{i-1})(t_{i+1}-t_{i})}
    &+\frac{-2x^*+t_{i}+t_{i+2}}{(t_{i+2}-t_{i})(t_{i+1}-t_{i})}=0\\
    (-2x^*+t_{i+1}+t_{i-1})(t_{i+2}-t_{i})&+(-2x^*+t_{i}+t_{i+2})(t_{i+1}-t_{i-1})=0\\
    x^*&=\frac{t_{i+1}t_{i+2}-t_{i}t_{i-1}}{t_{i+2}-t_{i}+t_{i+1}-t_{i-1}}
\end{align*}

$\bullet$ Because when $x < x^*$, $\frac{d}{dx}B_{i}^{2}(x) > 0$; when $x > x^*$, 
$\frac{d}{dx}B_{i}^{2}(x) < 0$. So $B_{i}^{2}(x)$ is at its maximum at $x^*$. Plug $x^*$
into $B_{i}^{2}(x)$. $B_{i}^{2}(x^*) < 1$ and $B_{i}^{2}(x) \geq 0$. So $B_{i}^{2}(x) \in [0,1)$.

$\bullet$
\begin{figure}[htbp]
    \centering
    \includegraphics[scale=0.7]{five_e.png}
    \caption{the figure of $B_{i}^{2}$ when $t_{i}=i$}
\end{figure}

\section{}
{\bfseries Solution:}\\
Because 
\begin{equation*}
    B_{i}^{0}(x)=\left\{\begin{aligned}
        &1 &if\quad x\in(t_{i-1},t_{i}]\\
        &0 &otherwise\\
    \end{aligned}\right.
\end{equation*}
So $B_{i}^{0}(x)=(t_{i}-x)_{+}^{0}-(t_{i-1}-x)_{+}^{0}$. From the Leibniz formula, we have
\begin{equation*}
    [t_{i-1},\ldots,t_{i+n}](t-x)_{+}^{n+1}=(t_{i-1}-x)[t_{i-1},\ldots,t_{i+n}](t-x)_{+}^{n}
    +[t_{i},\ldots,t_{i+n}](t-x)_{+}^{n}
\end{equation*}
Let $B_{i}^{1}(x)=\beta_{1}+\gamma_{1}$. Then
\begin{align*}
    \beta_{1}&=\frac{x-t_{i-1}}{t_{i}-t_{i-1}}B_{i}^{0}(x)\\
    &=\frac{x-t_{i-1}}{t_{i}-t_{i-1}}[(t_{i}-x)_{+}^{0}-(t_{i-1}-x)_{+}^{0}]\\
    &=(x-t_{i-1})[t_{i-1},t_{i}](t-x)_{+}^{0}\\
    &=[t_{i},t_{i}](t-x)_{+}^{0}-[t_{i-1},t_{i}](t-x)_{+}^{0}\\
    \gamma_{1}&=\frac{t_{i+1}-x}{t_{i+1}-t_{i}}B_{i}^{0}(x)\\
    &=(t_{i+1}-t_{i})[t_{i},t_{i+1}](t-x)_{+}^{0}\\
    &=(t_{i+1}-t_{i})[t_{i},t_{i+1}](t-x)_{+}^{0}+(t_{i}-x)[t_{i},t_{i+1}](t-x)_{+}^{0}\\
    &=[t_{i},t_{i+1}](t-x)_{+}^{1}-[t_{i},t_{i}](t-x)_{+}^{1}
\end{align*}
So $B_{i}^{1}(x)=[t_{i},t_{i+1}](t-x)_{+}^{1}-[t_{i-1},t_{i}](t-x)_{+}^{1}=(t_{i+1}-t_{i-1})
[t_{i-1},t_{i},t_{i+1}](t-x)_{+}^{1}$. And the same thing goes on. We can get
\begin{equation*}
    B_{i}^{2}=(t_{i+2}-t_{i-1})[t_{i-1},t_{i},t_{i+1},t_{i+2}](t-x)_{+}^{2}
\end{equation*}

\section{}
{\bfseries Solution:}\\
Because the B-spline is symmetric. So we can get $\int_{t_{i-1}}^{t_{i+n}} \,dB_{i}^{n}=0$.
From theorem 3.34, we have
\begin{equation*}
    \frac{d}{dx}B_{i}^{n}(x)=\frac{nB_{i}^{n-1}(x)}{t_{i+n-1}-t_{i-1}}-\frac{nB_{i+1}^{n-1}(x)}
    {t_{i+n}-t_{i}}
\end{equation*}
Then
\begin{align*}
    \int_{t_{i-1}}^{t_{i+n}} \frac{nB_{i}^{n-1}(x)}{t_{i+n-1}-t_{i-1}}-\frac{nB_{i+1}^{n-1}(x)}
    {t_{i+n}-t_{i}} \,dx = 0\\
    \int_{t_{i-1}}^{t_{i+n}} \frac{nB_{i}^{n-1}(x)}{t_{i+n-1}-t_{i-1}} \,dx=
    \int_{t_{i-1}}^{t_{i+n}} \frac{nB_{i+1}^{n-1}(x)}{t_{i+n}-t_{i}} \,dx\\
    \int_{t_{i-1}}^{t_{i+n}} \frac{B_{i}^{n-1}(x)}{t_{i+n-1}-t_{i-1}} \,dx=
    \int_{t_{i-1}}^{t_{i+n}} \frac{B_{i+1}^{n-1}(x)}{t_{i+n}-t_{i}} \,dx\\
\end{align*}
Because when $x\in (t_{i+n-1},t_{i+n})$, $B_{i}^{n-1}=0$ and when $x\in (t_{i-1},t_{i})$,
$B_{i+1}^{n-1}=0$. So 
\begin{align*}
    \int_{t_{i-1}}^{t_{i+n-1}} \frac{B_{i}^{n-1}(x)}{t_{i+n-1}-t_{i-1}} \,dx=
    \int_{t_{i}}^{t_{i+n}} \frac{B_{i+1}^{n-1}(x)}{t_{i+n}-t_{i}} \,dx
\end{align*}
Let $f(i-1)=\int_{t_{i-1}}^{t_{i+n-1}} \frac{B_{i}^{n-1}(x)}{t_{i+n-1}-t_{i-1}} \,dx$.
So we can get $f(i-1)=f(i)$. So $f(i)$ is constant and independent of its index i.

\section{}
{\bfseries Solution:}\\
$\bullet$ When m=4 and n=2, according to this theroem, $\tau_{2}(x_{i},x_{i+1},x_{i+2})
=[x_{i},x_{i+1},x_{i+2}]x^{4}$.\\
The LHS is
\begin{equation*}
    \tau_{2}(x_{i},x_{i+1},x_{i+2})=x_{i}x_{i+1}+x_{i}x_{i+2}+x_{i+1}x_{i+2}+x_{i}^{2}+
    x_{i+1}^{2}+x_{i+2}^{2}
\end{equation*}
Because we have
\begin{align*}
    &[x_{j}]x^{4}=x_{j}^{4}, (j=i,i+1,i+2)\\
    &[x_{j},x_{j+1}]x^{4}=\frac{x_{j+1}^{4}-x_{j}^{4}}{x_{j+1}-x_{j}}
    =(x_{j+1}+x_{j})(x_{j+1}^{2}+x_{j}^{2}), (j=i+1,i+2)
\end{align*}
The RHS is
\begin{align*}
    [x_{i},x_{i+1},x_{i+2}]x^{4}&=\frac{(x_{i+2}+x_{i+1})(x_{i+2}^{2}+x_{i+1}^{2})
    -(x_{i+1}+x_{i})(x_{i+1}^{2}+x_{i}^{2})}{x_{i+2}-x_{i}}\\
    &=\frac{(x_{i+2-x_{i}}-x_{i})x_{i+1}^{2}+(x_{i+2}^{2}-x_{i}^{2})x_{i+1}+
    x_{i+2}^{3}-x_{i}^{3}}{x_{i+2}-x_{i}}\\
    &=x_{i}x_{i+1}+x_{i}x_{i+2}+x_{i+1}x_{i+2}+x_{i}^{2}+
    x_{i+1}^{2}+x_{i+2}^{2}\\
    &=LHS
\end{align*}

$\bullet$ When n=0, $\tau_{m}(x_{i})=[x_{i}]x^{m}=x_{i}^{m}$ is obviously true.\\
Suppose when n=k(k=0,1,\ldots,m-1), that is also true.\\
Then when n=k+1, by Lemma 3.45, we have
\begin{align*}
    &(x_{i+k+1}-x_{i})\tau_{m-k-1}(x_{i},\ldots,x_{i+k+1})\\
    &=\tau_{m-k}(x_{i},\ldots,x_{i+k+1})-\tau_{m-k}(x_{i},\ldots,x_{i+k})-x_{i}\tau_{
    m-k-1}(x_{i},\ldots,x_{i+k+1})\\
    &=\tau_{m-k}(x_{i+1},\ldots,x_{i+k+1})+x_{i}\tau_{m-k-1}(x_{i},\ldots,x_{i+k+1})\\
    &-\tau_{m-k}(x_{i},\ldots,x_{i+k})-x_{i}\tau_{m-k-1}(x_{i},\ldots,x_{i+k+1})\\
    &=\tau_{m-k}(x_{i+1},\ldots,x_{i+k+1})-\tau_{m-k}(x_{i},\ldots,x_{i+k})\\
    &=[x_{i+1},\ldots,x_{i+k+1}]x^{m}-[x_{i},\ldots,x_{i+k}]x^{m}\\
    &=(x_{i+k+1}-x_{i})[x_{i},\ldots,x_{i+k+1}]x^{m}
\end{align*}
So $\tau_{m-k-1}(x_{i},\ldots,x_{i+k+1})=[x_{i},\ldots,x_{i+k+1}]x^{m}$. Which 
completes the proof.

\end{document}